As h->0, show that(explain why)
sin(x+h) = sinx +hcosx + o(h)
some thing like this always depends on what you already know and are allowed to use.
If you know that the derivative of sin x is cos x, then you can find the tangent line to sin x at $\displaystyle x= -x_0$:
$\displaystyle sin(x)= cos(x_0)(x- x_0)+ sin(x_0)$
Taking $\displaystyle x= x_0+ h$, $\displaystyle sin(x_0+ h)= cos(x_0)(h)+ sin(x_0)$ and, replacing $\displaystyle x_0$ with x,
sin(x+ h)= cos(x)h+ sin(x).
Another way would be to use the Taylor's series for sin(x) and cos(x):
$\displaystyle sin(x)= x- \frac{x^2}{2!}+ \frac{x^4}{4!}+ \cdot\cdot\cdot$
$\displaystyle cos(x)= 1- \frac{x^3}{3!}+ \frac{x^5}{5!}+ \dot\cdot\cdot$
Sorry person who is apparently not that awesome at maths.
"The above equation is wrong. The function on the LHS is the sine
function, whereas on the RHS you have a straight line. How can they
equal?
The tangent line is NOT equal to the curve. They only meet at ONE point:
the touching point."
Could someone else do my question?....
Do you know that if $\displaystyle \lim_{x\to a} f(x) = L$ then $\displaystyle f(x) = L + o(1)$ where the o(1) is as $\displaystyle x\to a$? Write down the difference quotient and use this to turn it into the tangent line estimate $\displaystyle f(x+h) = f(x) + f'(x) h +o(h)$, then set f(x) = sin(x).
Also, be nice.
edit: Or were you trying to prove that the derivative of sine is the cosine?